Smooth

**shilz222** · August 16th 2007, 01:03 PM

Let $f: \mathbb{R} \to \mathbb{R}$ be given. Assume that the square and cube of $f$ are smooth. Is $f$ smooth? That is , if $f \cdot f \in C^{\infty}$ and $f \cdot f \cdot f \in C^{\infty}$ , does it follow that $f \in C^{\infty}$ ?

**shilz222** · August 16th 2007, 01:17 PM

Or maybe we can put it this way: if $f^{n} \in C^{\infty}$ and $f^{n-1} \in C^{\infty}$ does it follow that $f \in C^{\infty}$ ?

**Rebesques** · August 17th 2007, 07:52 AM

Or maybe we can put it this way: if
$ f^{n} \in C^{\infty} $ and $ f^{n-1} \in C^{\infty} $ does it follow that $ f \in C^{\infty} $ ?

Not in general. You need the extra assumption that $f^n,f^{n-1}$ are never zero.

The proof is easy: Choose any natural number $m$ . Since $gcd(n,n-1)=1,$ there are integers $q, \ p$ such that $pn+q(n-1)=1$ . So $m=pmn+qm(n-1)$ ; Thus $f^m=(f^n)^{mp}(f^{n-1})^{qm}$ is differentiable as a product of differentiable functions. Note that the exponents $mp, \ mq$ may be negative, and so the assumption $f^n, f^{n-1}\neq0$ is essential.

**ThePerfectHacker** · August 17th 2007, 07:57 AM

I have another way to show this.

If $f^3\not = 0$ for any point and $f^3\in C^{\infty}$ then $(f^3)^{1/3} \in C^{\infty}$ thus $f\in C^{\infty}$ . Because the composition of infinitely differenciable functions is infinitely differenciable. Note, that $x^{1/3}$ is not differenciable at zero, which is why we need the condition that the function is never zero. However, I cannot come up with a conterexample.

This does not work with $f^2$ .

**topsquark** · August 17th 2007, 08:10 AM

Originally Posted by Rebesques

Not in general. You need the extra assumption that $f^n,f^{n-1}$ are never zero.

The proof is easy: Choose any natural number $m$ . Since $gcd(n,n-1)=1,$ there are integers $q, \ p$ such that $pn+q(n-1)=1$ . So $m=pmn+qm(n-1)$ ; Thus $f^m=(f^n)^{mp}(f^{n-1})^{qm}$ is differentiable as a product of differentiable functions. Note that the exponents $mp, \ mq$ may be negative, and so the assumption $f^n, f^{n-1}\neq0$ is essential.

Ah! So my thought that we may construct the function f as $\frac{f^3}{f^2}$ is at least not out of the ballpark. (This, of course, gives the same limitation that $f \neq 0$ at any point in its domain.) But this can't be the whole story because if we have
$f^2(x) = (x - 1)^4$
and
$f^3(x) = (x - 1)^6$

the function $f(x) = (x - 1)^2$ is, in fact, $C^{\infty}$ .

How might we get around this? (Or am I barking up the wrong tree again?

)

-Dan

**Rebesques** · August 17th 2007, 08:17 AM

Topsq, you have chosen a polynomial

Powers of that are always differentiable!

ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P

**topsquark** · August 17th 2007, 08:21 AM

Originally Posted by Rebesques

Topsq, you have chosen a polynomial

Powers of that are always differentiable!

ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P

But my $f^3$ and $f^2$ are $C^{\infty}$ and there is a point in the domain of both that are 0. According to you and TPH my f should not be $C^{\infty}$ . Or should we all be saying that we can't guarantee that such a $C^{\infty}$ f is possible under these conditions?

-Dan

**Rebesques** · August 17th 2007, 08:26 AM

According to you and TPH my f should not be...

Alright you win, add "polynomials and no-denominator ever functions out"

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